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Why y is a function of x ?

by mahmoud2011
Tags: function
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mahmoud2011
#19
Oct6-11, 04:00 AM
P: 88
Quote Quote by Fredrik View Post
The proof goes like this:

Let t0 be the real number such that [itex]x(t_0)=6[/itex]. By assumption, y satisfies [itex]y(t)=\sqrt{100-x^2(t)}[/itex] for all t in some open interval that contains t0. Denote the following functions by f,g and h respectively. [tex]
\begin{align}
& s\mapsto s^2\\
& s\mapsto 100-s\\
& s\mapsto \sqrt{s}
\end{align}[/tex] Since
[itex]y=h\circ g\circ f\circ x,[/itex]
x is differentiable at t0,
f is differentiable at x(t0),
g is differentiable at f(x(t0)),
h is differentiable at g(f(x(t0))).
the chain rule tells us that y is differentiable at t0.
yes , I know this proof but some it becomes very hard to solve the equation , for general how we can know that the equation is define y implicitly as a differentiable function of x
, is it good to me to take advanced course in Calculus to know how .

Thanks
Fredrik
#20
Oct6-11, 04:48 AM
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Quote Quote by mahmoud2011 View Post
yes , I know this proof but some it becomes very hard to solve the equation , for general how we can know that the equation is define y implicitly as a differentiable function of x
, is it good to me to take advanced course in Calculus to know how .

Thanks
Check out the implicit function theorem (the indented statement at the end of the section "statement of the theorem").
mahmoud2011
#21
Oct6-11, 06:37 AM
P: 88
so must I apply the the theorem for every time I use implicit differentiation ?


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